What is the area of the triangle inscribed in the parabola $y^2 = 4x$ whose vertices have $y$-coordinates $1, 2,$ and $4$?

$A$ normal with slope $\frac{1}{\sqrt{6}}$ is drawn from the point $(0, -\alpha)$ to the parabola $x^2 = -4ay$,where $a > 0$. Let $L$ be the line passing through $(0, -\alpha)$ and parallel to the directrix of the parabola. Suppose that $L$ intersects the parabola at two points $A$ and $B$. Let $r$ denote the length of the latus rectum and $s$ denote the square of the length of the line segment $AB$. If $r : s = 1 : 16$,then the value of $24a$ is. . . .

The shortest distance between the line $y = x$ and the curve $y^2 = x - 2$ is

The ends of the latus rectum of the parabola $x^2 + 8y = 0$ are

The smallest value of $x^2 - 3x + 3$ in the interval $(-3, 3/2)$ is

If $(2, k)$ is a point on the parabola passing through the points $(1, -3), (-1, 5), (0, 2)$ and having its axis parallel to the $Y$-axis,then $k$ is equal to